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3 years ago

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Chemistry

1 answer:

Veronika [31]3 years ago

7 0

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

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Answer:

E) Intramolecular bond angles change

Explanation:

Infrared Radiation:

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When IR radiation strike the molecule it absorbed by this molecule.

This radiation used to identify and study chemicals.

Infrared radiation interact with intra-bonds of the molecule.

Bonds in the molecules have vibrational translational and rotational movements
Due to these vibration, rotation and translation movement it absorb a radiation of specific frequency and wavelength
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So when Infrared light or radiation absorbed the intra-bonds of the molecule get affected and angles of these bonds changes.
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The correct option is option E

E) Intramolecular bond angles change

* Note:

it couldn't be option A as the frequency of IR is not enough to rotate a whole molecule

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It also not could be the option C as for the excitation of electrons require much higher energy.

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